Question

Three persons A, B and C at the corners of an equilateral triangle of side $X$ move at a constant speed $V$. Each person maintains a direction towards the person at the next corner. The time, the persons take to meet each other is :

• A. $\frac{2X}{3V}$
• B. $\frac{2X}{V}$
• C. $\frac{X}{V}$
• D. $\frac{4X}{3V}$

Answer 1

The correct option is A $\frac{2X}{3V}$

When A starts moving towards B, B has already started moving towards C and as B starts moving towards C, C has already started moving towards A, and as C starts moving towards A , A has already started moving towards B.

velocity of A is V along AB

the velocity of B is along BC . its component along BA is $Vcos60=\frac{V}{2}$. thus the separation AB decreases at the rate $V_{res}=V+\frac{V}{2}=\frac{3V}{2}$

since this rate is constant. the time taken in reducing the separation AB from $X$to $0$ is

$time=\frac{distance}{speed}=\frac{X}{\frac{3V}{2}}=\frac{2X}{3V}$